Question

The standard potentials are given as: 

\[\ce{E{^{\circ}_{Cu^{2+}/Cu}}}\] = +0.34 V, and \[\ce{E{^{\circ}_{Ag^{+}/Ag}}}\] = +0.80 V.

Calculate the cell potential (E) for the cell containing 0.100 M Ag⁺ and 4.00 M Cu²⁺ at 298 K.

Answer 1

Given: \[\ce{E{^{\circ}_{Cu^{2+}/Cu} = +0.34 V}}\]

\[\ce{E{^{\circ}_{Ag^{+}/Ag} = +0.80 V}}\]

We know that 

\[\ce{E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}}\]

= 0.80 − 0.34

= 0.46 V

By using the Nernst equation

`E = E_"cell"^circ - 0.0591/n log ((Cu^(2+))/((0.100)^2))`

Where n = 2, Cu²⁺ = 4.00 M, Ag⁺ = 0.100 M

∴ E = `0.46 - 0.0591/2 log (4.00/(0.100)^2)`

= 0.46 − 0.02955 log(400)

= 0.46 − 0.02955 × 2.602

= 0.46 − 0.0769

= 0.3831 V

∴ The cell potential is approximately 0.383 V.